FUNCTIONS OF TWO VARIABLES
Maths21a, O. Knill
FUNCTIONS, DOMAIN AND RANGE. We deal with functions f (x, y) of two variables defined on a domain
D in the plane. The p
domain is usually the entire plane like for f (x, y) = x2 + sin(xy). But there are cases
like when f (x, y) = 1/ 1 − (x2 + y 2 ), where the domain is a subset of the plane. The range of f is the set of
possible values of f .
EXAMPLES.
function f (x, y)
f (x, y) = sin(3x + 3y) − log(1 − x2 − y 2 )
f (x, y) = fp(x, y) = x2 + y 3 − xy + cos(xy)
f (x, y) = 4 − x2 − 2y 2
f (x, y) = 1/(x2 + y 2 − 1)
f (x, y) = 1/(x2 + y 2 )2
domain D
open unit disc x2 + y 2 < 1
entire plane R2
closed elliptic region x2 + 2y 2 ≤ 4
everything but the unit circle
everything but the origin
range f (D)
[−1, ∞)
entire real line
[0, 2]
entire real line
positive real axis
LEVEL SURFACES. We will later see surfaces which
are the three dimensional analog of level curves. if
f (x, y, z) is a function of three variables and c is a
constant then f (x, y, z) = c is a surface in space.
It is called a contour surface or a level surface.
For example if f (x, y, z) = 4x2 + 3y 2 + z 2 then the
contour surfaces are ellipsoids. We will see that in
the next lecture.
LEVEL CURVES
If f (x, y) is a function of two variables, then f (x, y) =
c = const is a curve or a collection of curves in the
plane. It is called contour curve or level curve.
For example, f (x, y) = 4x2 + 3y 2 = 1 is an ellipse.
Level curves allow to visualize functions of two variables f (x, y).
EXAMPLE. Let f (x, y) = x2 − y 2 . The set x2 − y 2 = 0 is the union of the sets x = y and x = −y. The set
x2 − y 2 = 1 consists of two hyperbola with with their tips at (−1, 0) and (1, 0). The set x2 − y 2 = −1 consists
of two hyperbola with their tips at (0, ±1).
EXAMPLE. f (x, y) = 1 − 2x2 −
y 2 . The contour curves f (x, y) =
1 − 2x2 + y 2 = c are the ellipses
2x2 + y 2 = 1 − c for c < 1.
SPECIAL LINES. Level curves are encountered every day:
Isobars:
Isoclines:
pressure
direction
Isothermes:
Isoheight:
temperature
height
For example, the isobars to the right show the lines of constant temperature in the north east of the US.
A SADDLE. f (x, y) = (x2 −
2
2
y 2 )e−x −y . We can here no more
find explicit formulas for the con2
2
tour curves (x2 − y 2 )e−x −y = c.
Lets try our best:
• f (x, y) = 0 means x2 − y 2 = 0 so that x = y, x = −y are contour curves.
CONTOUR MAP. Drawing several contour curves {f (x, y) = c} or several produces what one calls a contour
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• On y = ax the function is g(x) = (1 − a2 )x2 e−(1+a
)x2
.
• Because f (x, y) = f (−x, y) = f (x, −y), the function is symmetric with respect to reflections at the x and
y axis.
map.
The example shows the graph of the function f (x, y) = sin(xy). We draw the contour map of f : The curve
sin(xy) = c is xy = C, where C = arcsin(c) is a constant. The curves y = C/x are hyperbolas except for C = 0,
where y = 0 is a line. Also the line x = 0 is a contour curve. The contour map is a family of hyperbolas and
the coordinate axis.
TOPOGRAPHY. Topographical maps often show the curves of equal height. With the contour curves as
information, it is usually possible to get a good picture of the situation.
A SOMBRERO.
p The surface z =
f (x, y) = sin( x2 + y 2 ) has circles as contour lines.
ABOUT CONTINUITY. In reality, one sometimes has to deal with functions which are not smooth or not
continuous: For example, when plotting the temperature of water in relation to pressure and volume, one
experiences phase transitions, an other example are water waves breaking in the ocean. Mathematicians
have also tried to explain ”catastrophic” events mathematically with a theory called ”catastrophe theory”.
Discontinuous things are useful (for example in switches), or not so useful (for example, if something breaks).
DEFINITION. A function f (x, y) is continuous at (a, b) if f (a, b) is finite and lim(x,y)→(a,b) f (x, y) = f (a, b).
The later means that that along any curve ~r(t) with r(0) = (a, b), we have limt→0 f (~r(t)) = f (a, b).
Continuity for functions of more variables is defined in the same way.
EXAMPLE. f (x, y) = (xy)/(x2 + y 2 ). Because
lim(x,x)→(0,0) f (x, x) = limx→0 x2 /(2x2 ) = 1/2 and
lim(x,0)→(0,0) f (0, x) = lim(x,0)→(0,0) 0 = 0. The
function is not continuous.
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EXAMPLE. f (x, y) = (x2 y)/(x2 + y 2 ). In polar
coordinates this is f (r, θ) = r3 cos2 (θ) sin(θ)/r2 =
r cos2 (θ) sin(θ). We see that f (r, θ) → 0 uniformly
if r → 0. The function is continuous.
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